[tex]y'=(3x^2-3)f'(x^3-3x+m)\\y'=0\Leftrightarrow \left[\begin{array}{l} x= \pm 1\\x^3-3x=-3-m\\x^3-3x=3-m \end{array}\right.[/tex]
KS hàm $f(x)=x^3-3x$ có:

Vậy để hàm số có 4 cực trị thì:
[tex]\left[\begin{array}{l} 3-m \leq -2\\-3-m \geq 2 \\ -3-m=-2 \\ 3-m=2 \\\left\{\begin{array}{l}
3-m>2\\ -3-m<-2
\end{array}\right. \end{array}\right.\Leftrightarrow \left[\begin{array}{l} m \geq 5\\m \leq -5 \\ m=1 \\m=-1 \\
-1<m<1
\end{array}\right.[/tex]
Vậy chọn B